A square loop of side 10 cm and resistance 1 Omega is moved towards right with a constant velocity V_0 as shown in . The left arm of the loop is in a uniform magnetic field of 2 T. the field is perpendicular to the plane of the drawing and is going into it. the loop is connected to a network of resistors each of value 3 Omega with what speed should the loop be moved so that a steady current of 1 mA flows in the loop.

Explanation

The equivlant resistance of the network of the resistors, between P and Q will be 3 Omega
The totla resistance of the circuit is 1Omega + 3 Omega = 4 Omega
The emf induced in the loop is `epsilon = vBl = v_o (2T)(10cm)
The current in the loop will be i = (epsilon)/(R ) or, 1X10 ^(-3) A = (v_0(2T)(0
1 m)/(4 Omega) giving v_0 =( 4 Omega)(1X10^(-3) A)/(0
2 Tm) = 2 cm s^(-1)